Verify my proof: $\cup (\mathcal F \cap \mathcal G) \subseteq (\cup \mathcal F) \cap (\cup \mathcal G) $ I'm self learning from book "How to Prove it" by Velleman (3rd edition). I don't have access to a math professor, so I need a little help from the community. Please verify my proof.
Problem 18 pag. 140

Suppose $\mathcal F$ and $\mathcal G$ are families of sets. Prove that $\cup (\mathcal F \cap \mathcal G) \subseteq (\cup   \mathcal F) \cap (\cup \mathcal G) $

Proof. Let's introduce the following notations $A = \cup (\mathcal F \cap \mathcal G) $ and $ B = (\cup   \mathcal F) \cap (\cup \mathcal G) $. We must prove that $A \subseteq B$, this means that if we take an arbitrary element $x$ from $A$ it must be also an element from $B$. $x \in A$ means that there is a set $z$ which is a member of both families of sets $\mathcal F$ and $\mathcal G$, so $z \in \mathcal F$ and $z \in \mathcal G$ and $x \in z$. Let's mark this sentence as (1).
$x \in B$ means that there is a set $v$ which is a member of $\mathcal F$ and a set $w$ which is a member of $\mathcal G$ so $v \in \mathcal F$ and $w \in \mathcal G$ and $x \in v \cap w$. Let's mark this sentence as (2). Now observe the difference from (1), $z$ is a member of both $\mathcal F$ and $\mathcal G$, but $v \in \mathcal F$ and $w \in \mathcal G$, as a particular case $v=w=z$. But generally speaking elements from $A$ are a subset of elements from $B$ $\blacksquare $
 A: From the logic, you are completely right.
Not that I want to suggest writing too much formalism (on the contrary, proofs drowning in "$\Rightarrow$" and "$\therefore$" and such are awful to read!), but for my personal taste, this is  too wordy (in particularly those "let's mark this sentence ..." parts, and the "$v=w=z$" may even be confusing) -
Compare with this (just a style suggestion):

Proof. Let $x$ be an arbitrary element of $\bigcup (\mathcal F\cap \mathcal G)$. Then by definition of $\bigcup$, there exists $y\in \mathcal F\cap \mathcal G$ with $x\in y$. By definition of $\cap$,  we have both $y\in\mathcal F$ and $y\in\mathcal G$. But $x\in y\in\mathcal F$ implies $x\in\bigcup F$, and $x\in y\in\mathcal G$ implies $x\in\bigcup G$. From $x\in\bigcup F$ and $x\in \bigcup\mathcal G$, we get $x\in (\bigcup \mathcal F)\cap (\bigcup \mathcal G)$.
So in summary, $x\in \bigcup (\mathcal F\cap \mathcal G)$ implies $x\in (\bigcup \mathcal F)\cap (\bigcup \mathcal G)$, in other words, $\bigcup (\mathcal F\cap \mathcal G)\subseteq (\bigcup \mathcal F)\cap (\bigcup \mathcal G)$, as was to be shown. $\square$

